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On 8/4/2024 1:22 AM, Fred. Zwarts wrote:But the "DDD emulated by HHH" is the program DDD above, that calls that HHH. If that HHH aborts its emulation and returns, the program, that is the DDD that was emulated by HHH, does reach the return.Op 03.aug.2024 om 18:35 schreef olcott:>>>> ∞ instructions of DDD correctly emulated by HHH[∞] nevervoid DDD()reach their own "return" instruction final state.>
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So you are saying that the infinite one does?
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Dreaming again of HHH that does not abort? Dreams are no substitute for facts.
The HHH that aborts and halts, halts. A tautology.
{
HHH(DDD);
return;
}
That is the right answer to the wrong question.
I am asking whether or not DDD emulated by HHH
reaches its "return" instruction.
No, it must EMULATE that HHH, (not produce the output that HHH generates, but emulate the process of getting that output).The correct simulation of a halting program halts. A truism.So by correctly you must mean that HHH incorrectly
HHH cannot possibly simulate itself correctly.
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skips the call from DDD to HHH(DDD).
But only FINITE recursion if HHH will ever abort its emulation, and thus is not non-halting unless HHH just never answers, so it isn't the decider you claim.I am happy that you were not a member of our team when we developed simulators to check the design of big detector systems.Crash dummy simulators?
Not x86 emulators.
We knew that a simulation is only correct if it matches the reality. But you seem to think that it is correct that a simulator does not match the reality.The reality is that DDD does call HHH(DDD) in recursive
simulation. When you ignore this then you are out-of-touch
with reality.
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